Use the localized electron model to describe the bonding in \(\mathrm{H}_{2} \mathrm{CO}\) (carbon is the central atom).

Short Answer

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In H2CO, the central carbon atom exhibits sp2 hybridization and forms a trigonal planar molecular geometry. Carbon forms sigma bonds with two hydrogen atoms and one oxygen atom, while the double bond between carbon and oxygen consists of one sigma bond and one pi bond. The bonding in H2CO can be described using the localized electron model, with carbon's three sp2 hybrid orbitals forming sigma bonds and the pi bond arising from the overlap of unhybridized p orbitals in carbon and oxygen.

Step by step solution

01

Determine the lewis structure of H2CO

To draw the Lewis structure of H2CO, we first need to count the total number of valence electrons for each atom. Hydrogen has 1 valence electron, Carbon has 4 valence electrons, and Oxygen has 6 valence electrons. The molecule's total valence electrons are: 2 (Hydrogen atoms) x 1 (valence electrons each) + 1 (Carbon atom) x 4 (valence electrons) + 1 (Oxygen atom) x 6 (valence electrons) = 12 valence electrons. We will use the total valence electrons to form bonds between the atoms: 1. Place the least electronegative atom, Carbon, in the center. 2. Attach the other atoms around the central atom with single bonds. 3. Distribute the remaining valence electrons as lone pairs to complete the octet rule for all atoms except hydrogen, which only needs 2 electrons. The Lewis structure of H2CO is: O || C H H
02

Identify the hybridization of the central carbon atom

To identify the hybridization of the central carbon atom, we will count the number of electron groups (lone pairs and bonding atoms) surrounding it. In the case of H2CO, there are 3 electron groups surrounding the carbon atom: 1. A double bond with oxygen (1 group as each double bond is considered as one electron group) 2. Two single bonds with two hydrogen atoms. Three electron groups lead to sp2 hybridization.
03

Determine the molecular geometry of the molecule based on the hybridization

Based on the sp2 hybridization, the molecular geometry of H2CO will be trigonal planar.
04

Describe the bonding according to the electron localization model

Using the localized electron model, we can describe the bonding in H2CO as follows: 1. The central carbon atom forms three sp2 hybrid orbitals, which are used to form sigma bonds with two hydrogen atoms and one oxygen atom. 2. The oxygen atom forms a sigma (single) bond with the carbon atom. Oxygen also has two lone pairs of electrons in its remaining two p orbitals. 3. The double bond between carbon and oxygen consists of one sigma bond (from the overlap of sp2 orbitals) and one pi bond (from the overlap of unhybridized p orbitals in both carbon and oxygen). In summary, H2CO has a trigonal planar molecular geometry with a carbon central atom in an sp2 hybridization state. The carbon atom forms sigma bonds with hydrogen and oxygen atoms, and a pi bond with the oxygen atom.

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Most popular questions from this chapter

Why are \(d\) orbitals sometimes used to form hybrid orbitals? Which period of elements does not use \(d\) orbitals for hybridization? If necessary, which \(d\) orbitals \((3 d, 4 d, 5 d\), or \(6 d)\) would sulfur use to form hybrid orbitals requiring \(d\) atomic orbitals? Answer the same question for arsenic and for iodine.

What are molecular orbitals? How do they compare with atomic orbitals? Can you tell by the shape of the bonding and antibonding orbitals which is lower in energy? Explain.

The \(\mathrm{N}_{2} \mathrm{O}\) molecule is linear and polar. a. On the basis of this experimental evidence, which arrangement, NNO or NON, is correct? Explain your answer. b. On the basis of your answer to part a, write the Lewis structure of \(\mathrm{N}_{2} \mathrm{O}\) (including resonance forms). Give the formal charge on each atom and the hybridization of the central atom. c. How would the multiple bonding in \(\mathrm{N} \equiv \mathrm{N}-\ddot{\mathrm{O}}\) : be described in terms of orbitals?

Cyanamide \(\left(\mathrm{H}_{2} \mathrm{NCN}\right)\), an important industrial chemical, is produced by the following steps: $$ \begin{aligned} \mathrm{CaC}_{2}+\mathrm{N}_{2} & \longrightarrow \mathrm{CaNCN}+\mathrm{C} \\\ \mathrm{CaNCN} & \stackrel{\text { Acid }}{\longrightarrow} \mathrm{H}_{2} \mathrm{NCN} \\ \text { Cyanamide } \end{aligned} $$ Calcium cyanamide (CaNCN) is used as a direct-application fertilizer, weed killer, and cotton defoliant. It is also used to make cyanamide, dicyandiamide, and melamine plastics: Dicyandiamide not shown) a. Write Lewis structures for \(\mathrm{NCN}^{2-}, \mathrm{H}_{2} \mathrm{NCN}\), dicyandiamide, and melamine, including resonance structures where appropriate. b. Give the hybridization of the \(\mathrm{C}\) and \(\mathrm{N}\) atoms in each species. c. How many \(\sigma\) bonds and how many \(\pi\) bonds are in each species? d. Is the ring in melamine planar? e. There are three different \(\mathrm{C}-\mathrm{N}\) bond distances in dicyandiamide, \(\mathrm{NCNC}\left(\mathrm{NH}_{2}\right)_{2}\), and the molecule is nonlinear. Of all the resonance structures you drew for this molecule, predict which should be the most important.

Compare and contrast bonding molecular orbitals with antibonding molecular orbitals.

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